我假设你的列是数字的。

import datetime as dt

df = pd.DataFrame({'code': [19501, 19502, 19711, 73048]})
df['days'] = pd.to_timedelta(df['code']//100, 'D')
df['half-hours'] = df['code']%100
df['hours'] = pd.to_timedelta(df['half-hours']//2, 'h')
df['minutes'] = pd.to_timedelta(df['half-hours']%2*30, 'm')

base_day = dt.datetime(2009, 1, 1) - dt.timedelta(days = 195)

df['dt0'] = base_day + df.days + df.hours + df.minutes - dt.timedelta(minutes = 30)
df['dt1'] = base_day + df.days + df.hours + df.minutes - dt.timedelta(seconds = 1)

一个简单的解决方案，添加的天2008-06-20，添加的（时间-1）* 30 min;

df = pd.DataFrame({'code': [19501, 19502, 19711, 73048]})

d, t = df['code'].divmod(100)

df['datetime'] = (
   pd.to_timedelta(d, unit='D')
     .add(pd.Timestamp('2008-06-20'))
     .add(pd.to_timedelta((t-1)*30, unit='T'))
)

• 注意：这是周期的开始，对于结束，请将(t-1)*30替换为t*30-1。*

输出：

code            datetime
0  19501 2009-01-01 00:00:00
1  19502 2009-01-01 00:30:00
2  19711 2009-01-03 05:00:00
3  73048 2010-06-20 23:30:00